Why is Quantum Field Theory Local?

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In summary: I am using... think that entanglement means "nonlocal". Quantum Field Theory includes entanglement, because it includes non-relativistic QM as a special case and makes all of the same predictions for that case.
  • #1
LarryS
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Why is Quantum Field Theory local?
Under the Schrodinger Picture, nonrelativistic Quantum Mechanics for a fixed number of particles is highly nonlocal, e.g. Quantum Entanglement.

But Quantum Field Theory is local. Why is that? Is it because QFT was created to accommodate SR, which, as a classical theory, is local?

As always, thanks in advance.
 
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  • #2
referframe said:
Quantum Field Theory is local.

Not in the sense you are using the term "local", since you are saying that entanglement means "nonlocal", and QFT includes entanglement, since it includes non-relativistic QM as a special case and makes all of the same predictions for that case.

QFT is "local" in the sense that spacelike separated measurements, including those on entangled particles, must commute--their results must not depend on the order in which they are made (since the ordering of spacelike separated measurements is not invariant). But this is a different meaning of "local" from the one you are using (your meaning of "local" is basically "correlations satisfy the Bell inequalities", which is violated by measurements on entangled quantum particles).
 
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  • #3
Another sense in which QFT is local is that its Hamiltonian (or Lagrangian) is local.
 
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  • #4
Demystifier said:
Another sense in which QFT is local is that its Hamiltonian (or Lagrangian) is local.
What's a local Hamiltonian?
 
  • #5
martinbn said:
What's a local Hamiltonian?
Roughly, a Hamiltonian that can be written in the form
$$H=\int d^3x\,{\cal H}(x)$$
where ##{\cal H}(x)## is Hamiltonian density. It's not perfectly precise, but I'll not nitpick unless you insist.
 
  • #6
Demystifier said:
Roughly, a Hamiltonian that can be written in the form
$$H=\int d^3x\,{\cal H}(x)$$
where ##{\cal H}(x)## is Hamiltonian density. It's not perfectly precise, but I'll not nitpick unless you insist.
Is it easy to write an example of a local and of a nonloval Hamiltonian?
 
  • #7
martinbn said:
Is it easy to write an example of a local and of a nonloval Hamiltonian?
Yes.

Local:
$$H=\int d^3x\, \frac{1}{2}\pi(x)\pi(x)$$
where ##\pi(x)## is canonical momentum density associated with the field ##\phi(x)##. (For simplicity I gave an example which does not have a ##\phi##-dependent term and does not describe a Lorentz covariant theory. Contrary to a widespread prejudice, locality and Lorentz covariance are independent requirements.)

Nonlocal:
$$H=\int d^3x\int d^3x' \, \frac{1}{2}\pi(x)G(x,x')\pi(x')$$
where ##G(x,x')## is not proportional to ##\delta^3(x-x')##. For example,
$$G(x,x')=\frac{const}{|x-x'|}$$
 
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  • #8
Demystifier said:
Yes.

Local:
$$H=\int d^3x\, \frac{1}{2}\pi(x)\pi(x)$$
where ##\pi(x)## is canonical momentum density associated with the field ##\phi(x)##. (For simplicity I gave an example which does not have a ##\phi##-dependent term and does not describe a Lorentz covariant theory. Contrary to a widespread prejudice, locality and Lorentz covariance are independent requirements.)

Nonlocal:
$$H=\int d^3x\int d^3x' \, \frac{1}{2}\pi(x)G(x,x')\pi(x')$$
where ##G(x,x')## is not proportional to ##\delta^3(x-x')##. For example,
$$G(x,x')=\frac{const}{|x-x'|}$$
Well, they are both in the form $$H=\int d^3x\,\cal{H}(x)$$

I was looking for an example, where you can explicitly write down the operator.
 
  • #9
referframe said:
Summary:: Why is Quantum Field Theory local?

Under the Schrodinger Picture, nonrelativistic Quantum Mechanics for a fixed number of particles is highly nonlocal, e.g. Quantum Entanglement.

But Quantum Field Theory is local. Why is that? Is it because QFT was created to accommodate SR, which, as a classical theory, is local?

As always, thanks in advance.
As already stated by other answers in this thread the confusion reflected in this question is due to the different meaning different communities of scientists and philosophers put into the words "local" and "non-local".

I'm a high-energy /particlenuclear physicist, and for me relativistic QFT is "local", because it's constructed to be local. The meaning in this case is that all local observables, represented by field operators ##\hat{O}(t,\vec{x})## (in the Heisenberg picture, which is the most natural picture of time evolution to discuss this issue) commute with the energy-density operator ##\hat{\mathcal{H}}(t,\vec{x})## at space-like separated arguments, i.e.,
$$[\hat{\op{O}}(x),\hat{\op{\mathcal{H}}(y)]=0 \quad \text{if} \quad (x-y)^2<0,$$
where I'm using the west-coast convention with ##\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)##, such that four vectors are space-like separated if their Minkowski product with themselves is negative. This implies physically that no measurement done at a place can have an effect on any event (particularly also any other measurement) which is space-like separated from this measurement. This for sure realizes causality in the relativistic sense by construction, i.e., there's no way to communicate faster than light in any way. That's why it's also called microcausality condition.

It is usually realized by constructing field-operators that transform in the same local way as do the corresponding classical fields under Poincare transformations, fulfilling canonical equal-time commutation (bosons) or anti-commutation relations and then writing down a Lagrange density which is Poincare invariant and depends only on the fields and their first derivative at one spacetime point.

In addition it also ensures the unitarity of the S-matrix and the linked-cluster theorem (see Weinberg, QT of fields, vol. I for the details).

Of course also an in this sense local relativistic QFT necessarily implies the existence of entangled states, but this has nothing to do with "non-locality", but it rather describes "inseparability" (as Einstein put it), i.e., it implies correlations between far distant parts of a single (!) quantum system when measured at far-distant places.

[Edit: Erased interpretational statements which don't belong to this scientific part of the QM forum]

The confusion comes into the discussion, because some physicists and many philosophers name the long-range correlations due to entanglement (i.e., "inseparability") "non-locality", but that has of course a different meaning than the "locality of interactions/transformation properties of field operators under Poincare transformations" in relativistic local QFTs.
 
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  • #10
martinbn said:
Well, they are both in the form $$H=\int d^3x\,\cal{H}(x)$$
That's why I said that it was not perfectly precise. I should have said something like "... where ##{\cal H}(x)## only depends on fields and its derivatives at ##x##". But in the examples it should be clear what I meant.

martinbn said:
I was looking for an example, where you can explicitly write down the operator.
What do you mean by that, can you give an example of an "explicitly written down operator"?
 
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  • #11
vanhees71 said:
1. This in turn implies that the correlations cannot be due to the mutual influence of the measurements at far distant places when the measurement events are at space-like separated places in spacetime.

2. The confusion comes into the discussion, because some physicists and many philosophers name the long-range correlations due to entanglement (i.e., "inseparability") as "non-locality", but that has of course another meaning than the "locality of interactions" in relativistic local QFTs.

1. Again, this sentence represents an opinion is not shared by most physicists doing work on entanglement. Of course, this is not a popularity contest. The OP has a right to know, and I suspect he has followed discussions on this point previously from his question.

2. Virtually the entirety of the physics community calls the correlations evidence of "quantum non-locality". No one understands the mechanism enough to say with certainty anything is happening that is FTL; certainly I don't.

I won't quote the community or otherwise comment further because we have already been down this ridiculous rabbit hole too many times already. Bell tests demonstrate quantum non-locality, in fact that's the generally accepted definition... period.

Cheers,

-DrC
 
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  • #12
DrChinese said:
Virtually the entirety of the physics community calls the correlations evidence of "quantum non-locality".

The statement that "QFT is local" is also a very common statement, and, as has been posted previously in this thread, in order to reconcile this statement with the statement that Bell inequality violations mean "nonlocality", one has to recognize that the term "local" is being used in two different senses.
 
  • #13
Demystifier said:
What do you mean by that, can you give an example of an "explicitly written down operator"?
Your examples have integrals of ##\pi(x)##, for which you haven't given an explicit formula. As for examples take the Laplace operator, it is written down explicitly as ##\sum\frac{\partial^2}{\partial x_i^2}##.
 
  • #14
DrChinese said:
1. Again, this sentence represents an opinion is not shared by most physicists doing work on entanglement. Of course, this is not a popularity contest. The OP has a right to know, and I suspect he has followed discussions on this point previously from his question.

2. Virtually the entirety of the physics community calls the correlations evidence of "quantum non-locality". No one understands the mechanism enough to say with certainty anything is happening that is FTL; certainly I don't.

I won't quote the community or otherwise comment further because we have already been down this ridiculous rabbit hole too many times already. Bell tests demonstrate quantum non-locality, in fact that's the generally accepted definition... period.

Cheers,

-DrC
I said that the terms "local" or "non-local" are used with different meanings in different communities, and I told you what HEP physicist mean when they say "local relativistic QFT". This has to be distinguished from what the interpretation-of-quantum-theory community (a mixture of physicists and, unfortunately, also philosophers) understand under "local" and "non-local". It's not an opinion of either group but just using the same words for different meanings.

I prefer to call correlations correlations and use locality/non-locality only in the clear mathematical meaning of the HEP community. That's all.

All I can say is that one has to carefully figure out, what's meant by any specific writer when he says "local" or "non-local".

Whether you draw the logical conclusion about the causality when measuring far-distant entangled observables or not is of course a matter of opinion and interpretation. So I shouldn't have written this part. I'll erase the corresponding statements. It belongs to the interpretational section of these forums, I'm no longer interested to participate in.
 
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  • #15
Just one short statement: I still believe, of course, that the mathematical properties of local relativistic QFTs (local in the HEP sense) inevitably excludes any fast-than-like propagation of signals.
 
  • #16
vanhees71 said:
Just one short statement: I still believe, of course, that the mathematical properties of local relativistic QFTs (local in the HEP sense) inevitably excludes any fast-than-like propagation of signals.

An "agnostic" would be more tentative. As Peter Mittelstaedt remarks in “Quantum Holism, Superluminality, and Einstein Causality”: “Finally, we analyze these arguments and show that the micro-causality condition of relativistic quantum field theory excludes entanglement induced superluminal signals but that this condition is justified by the exclusion of superluminal signals. Hence, we are confronted here with a vicious circle, and the question whether there are superluminal EPR-signals cannot be answered in this way.
 
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  • #17
This is interpretation, i.e., not subject to this forum.

Physicswise the microcausality condition together with the locality of the unitary representations of the proper orthochronous Poincare group indeed guarantees (a) causality, (b) the unitarity and Poincare invariance of the S-matrix, and (c) the validity of the linked cluster principle. Thus the inevitable consequence indeed is that there are no faster-than light signals and no violation of causality within such a "local" relativistic QFT.

One should be aware, as Weinberg stresses in his book, that this is a sufficient but not necessary condition. So it might well be that there are causal "non-local" relativistic QFTs. I'm not aware, whether somebody has constructed such a thing and which empirical consequences such a modification would imply in comparison to the local relativistic QFT making up the Standard Model.

I don't know, where, from a physics point of view should there be a "vicious circle" in this (purely mathematical) arguments.
 
  • #18
vanhees71 said:
This is interpretation, i.e., not subject to this forum.

What are you saying is interpretation? It looks to me like both the Mittelstaedt remarks and your response to them are, as you say, purely mathematical arguments involving the math of QFT, and don't involve any choice of QM interpretation.
 
  • #19
I meant the statement that it's a vicious circle. From a mathematical point of view there's no circle. It's the statement of a sufficient condition fulfilling the physically plausible properties (a)-(c). There's no circular argument. You can of course argue that also this is interpretation :oldbiggrin:.
 
  • #20
Lord Jestocost said:
An "agnostic" would be more tentative. As Peter Mittelstaedt remarks in “Quantum Holism, Superluminality, and Einstein Causality”: “Finally, we analyze these arguments and show that the micro-causality condition of relativistic quantum field theory excludes entanglement induced superluminal signals but that this condition is justified by the exclusion of superluminal signals. Hence, we are confronted here with a vicious circle, and the question whether there are superluminal EPR-signals cannot be answered in this way.

The Mittelstaedt statement is erroneous. The microcausality condition of relativistic QFT includes entanglement induced superluminal signals. There is no vicious circle. Nonetheless, the entanglement induced superluminal signals do not permit superluminal signalling, where "signals" and "signalling" are used in two different senses.
 
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  • #21
For me it's very clear that the microcausality condition excludes superluminal signals. You have to distinguish between causal effects (interactions) and correlations described by the (entangled) quantum states. Your last sentence admits this, and it's just somewhat confusing to say there's superluminal signalling but no superluminal signal.

If you want to introduce some new notion, you have to clearly define it. So what do you understand under "superluminal signalling"? It can for sure not mean causal effects by interactions because of the microcausality feature. That's a mathematical fact and indeed no question of opinion or "interpretation".
 
  • #22
atyy said:
The Mittelstaedt statement is erroneous. The microcausality condition of relativistic QFT includes entanglement induced superluminal signals. There is no vicious circle. Nonetheless, the entanglement induced superluminal signals do not permit superluminal signalling, where "signals" and "signalling" are used in two different senses.
So what does "signal" mean?
 
  • #23
martinbn said:
So what does "signal" mean?
That the entangled partner particle collapses to a 100% correlated state. Signalling doesn't take place, is what he is saying. 'Signals' is also a misnomer as no signals are going from one entangled partner particle to the other.
Assuming classicality and realism before measurement leads to the well-known EPR problem.
 
  • #24
But collapsing cannot "signal" anything (despite the fact that the collapse assumption is highly problematic and fortunately unnecessary for the application of quantum theory as far as any physics is concerned).
 
  • #25
vanhees71 said:
I meant the statement that it's a vicious circle. From a mathematical point of view there's no circle. It's the statement of a sufficient condition fulfilling the physically plausible properties (a)-(c). There's no circular argument. You can of course argue that also this is interpretation

The point is: You can believe in your physical theory, that's up to you. However, as Kent A. Peacock puts it in his thesis “Peaceful Coexistence or Armed Truce? Quantum Nonlocality and the Spacetime View of the World” (1991):

….. that the proof of a result on the basis of a theory which was ‘constructed to ensure’ that result is no proof at all. As I have noted, all the proof shows is that the construction was successful: it accomplishes what it was intended to accomplish.
 
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  • #26
It's no proof of course. All that counts is whether it stands stringent empirical tests.
 
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  • #27
martinbn said:
Your examples have integrals of ##\pi(x)##, for which you haven't given an explicit formula. As for examples take the Laplace operator, it is written down explicitly as ##\sum\frac{\partial^2}{\partial x_i^2}##.
In that sense, we can represent it by a functional derivative
$$\pi(x)=-i\hbar\frac{\delta}{\delta\phi(x)}$$
 
  • #28
Demystifier said:
In that sense, we can represent it by a functional derivative
$$\pi(x)=-i\hbar\frac{\delta}{\delta\phi(x)}$$
And what is ##\phi##? And what is the integral equal to? I just want to see e very specific example.
 
  • #29
martinbn said:
And what is ##\phi##? And what is the integral equal to? I just want to see e very specific example.
If you want a crash course in functional calculus for physicists, please start a separate thread!
 
  • #30
Demystifier said:
If you want a crash course in functional calculus for physicists, please start a separate thread!
No, I just want an operator.
 
  • #35
martinbn said:
Thanks!

Just point out that this is yet another sense in which local/nonlocal is used. For example all differential operators are local, in this way, so Newtonian gravity can be considered local, because the Laplace operator that appears in the Poisson equation is local.
Never thought that way, but yes, in that sense Newtonian gravity is local.
 
<h2>1. What is Quantum Field Theory (QFT)?</h2><p>Quantum Field Theory is a theoretical framework used to describe the behavior of particles and fields at the quantum level. It combines principles from quantum mechanics and special relativity to explain the interactions between particles and their associated fields.</p><h2>2. What does it mean for QFT to be "local"?</h2><p>In QFT, "local" refers to the concept that the interactions between particles and fields are limited to a specific region of space and time. This means that the effects of a particle's interaction with a field can only be observed within a finite distance and time interval.</p><h2>3. Why is it important for QFT to be local?</h2><p>The principle of locality is essential for maintaining causality in physical systems. It ensures that the effects of an interaction cannot be observed before the interaction itself has occurred. Additionally, the concept of locality is crucial in the development of renormalization techniques in QFT, which are used to eliminate infinities in calculations.</p><h2>4. What evidence supports the idea that QFT is local?</h2><p>Experimental observations, such as the decay of unstable particles, have shown that the effects of interactions are limited to a specific region of space and time. Additionally, the mathematical framework of QFT, which is based on the principles of locality, has been successful in predicting and explaining a wide range of phenomena.</p><h2>5. Are there any theories that challenge the idea of QFT being local?</h2><p>There have been attempts to develop non-local versions of QFT, such as non-local quantum field theories and string theory. However, these theories have not yet been fully developed and lack experimental evidence to support them. Additionally, the principle of locality is deeply ingrained in the foundations of modern physics, making it a fundamental concept that is not easily challenged.</p>

1. What is Quantum Field Theory (QFT)?

Quantum Field Theory is a theoretical framework used to describe the behavior of particles and fields at the quantum level. It combines principles from quantum mechanics and special relativity to explain the interactions between particles and their associated fields.

2. What does it mean for QFT to be "local"?

In QFT, "local" refers to the concept that the interactions between particles and fields are limited to a specific region of space and time. This means that the effects of a particle's interaction with a field can only be observed within a finite distance and time interval.

3. Why is it important for QFT to be local?

The principle of locality is essential for maintaining causality in physical systems. It ensures that the effects of an interaction cannot be observed before the interaction itself has occurred. Additionally, the concept of locality is crucial in the development of renormalization techniques in QFT, which are used to eliminate infinities in calculations.

4. What evidence supports the idea that QFT is local?

Experimental observations, such as the decay of unstable particles, have shown that the effects of interactions are limited to a specific region of space and time. Additionally, the mathematical framework of QFT, which is based on the principles of locality, has been successful in predicting and explaining a wide range of phenomena.

5. Are there any theories that challenge the idea of QFT being local?

There have been attempts to develop non-local versions of QFT, such as non-local quantum field theories and string theory. However, these theories have not yet been fully developed and lack experimental evidence to support them. Additionally, the principle of locality is deeply ingrained in the foundations of modern physics, making it a fundamental concept that is not easily challenged.

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